[confusedstud]

My notes states that when the entropy change of the universe is 0 it is a reversible change while when the entropy change of the universe is greater than 0, it is an irreversible change.

So in the case where two heat reservoirs are placed side by side with one at 100K and the other at 200K, the total entropy change will be as follows:
-q/200K+q/100K>0 So in this case is this a reversible or irreversible change?

I would say that it is an irreversible change however, we are using a reversible change to calculate the entropy change of the universe. Is this correct?

[mjc123]

Since entropy is a state function, ΔS depends only on the initial and final states, not on the route taken between them. Therefore if you can find a reversible path, and calculate the heat flow and thence the entropy change, the entropy change of the system will be the same even if you take an irreversible path between the same two states.
Let's consider your scenario: two heat reservoirs which we'll call A and B, at temperatures of 100K and 200K respectively.
Suppose that, keeping A and B separate, we bring into contact with them two more heat reservoirs, C into contact with A and D with B, with temperatures 100+δ and 200-δ respectively, where δ is very small, so that heat flow is effectively reversible, and we allow a quantity of heat q to flow from C to A, and from B to D. Then we take away reservoirs C and D. The net result is that we have transferred q reversibly from B to A , with a reversible entropy change ΔS_system = q/100 - q/200 = q/200. As regards the surroundings, we have transferred q from C to D, with a change of entropy q/(200-δ) - q/(100+δ) ≈ -q/200*(1-5δ/200). We can make δ as small as we like, so as δ 0, ΔS_surr -q/200 and ΔS_uni  0.
Now consider your original procedure  transferring the heat from B to A directly. The initial and final states of the system (A and B) are the same as in the scenario above - heat q has been transferred from B to A. Therefore the entropy change of the system is the same, q/200. But this time we have done nothing to the surroundings, so ΔS_surr = 0 and ΔS_uni = q/200. So (to answer your question) this is an irreversible process, but the entropy change of the system (not the surroundings) is the same as for a reversible process.

[confusedstud]

Since entropy is a state function, ΔS depends only on the initial and final states, not on the route taken between them. Therefore if you can find a reversible path, and calculate the heat flow and thence the entropy change, the entropy change of the system will be the same even if you take an irreversible path between the same two states.
Let's consider your scenario: two heat reservoirs which we'll call A and B, at temperatures of 100K and 200K respectively.
Suppose that, keeping A and B separate, we bring into contact with them two more heat reservoirs, C into contact with A and D with B, with temperatures 100+δ and 200-δ respectively, where δ is very small, so that heat flow is effectively reversible, and we allow a quantity of heat q to flow from C to A, and from B to D. Then we take away reservoirs C and D. The net result is that we have transferred q reversibly from B to A , with a reversible entropy change ΔS_system = q/100 - q/200 = q/200. As regards the surroundings, we have transferred q from C to D, with a change of entropy q/(200-δ) - q/(100+δ) ≈ -q/200*(1-5δ/200). We can make δ as small as we like, so as δ 0, ΔS_surr -q/200 and ΔS_uni  0.
Now consider your original procedure  transferring the heat from B to A directly. The initial and final states of the system (A and B) are the same as in the scenario above - heat q has been transferred from B to A. Therefore the entropy change of the system is the same, q/200. But this time we have done nothing to the surroundings, so ΔS_surr = 0 and ΔS_uni = q/200. So (to answer your question) this is an irreversible process, but the entropy change of the system (not the surroundings) is the same as for a reversible process.

Hmm shouldn't we consider the ΔS_uni to be both -q/200+q/100 for both heat reservoirs? Where its -q for the 200K reservoir because it is loosing heat?

Because if we only considered the brick B would it be -q/200 instead of the q/200 you mentioned?

[mjc123]

I'm sorry, I don't understand your question.

[confusedstud]

I'm sorry, I don't understand your question.

Sorry I misread your first answer I understand it now. The system consists of both heat reservoirs so Delta S(system)=q/200 and the Delta S(surr)=0 as there is no heat loss or gain from the system (is this statement correct?). So delta S(universe)=q/200 which makes it an irreversible process using a reversible process calculation?

But I don't quite understand your first scenario where upon removal of C and D then its the same as transferring heat from A to B and why in this case Delta S(surr)=-q/200. Could you explain this part again?

[mjc123]

Does this make it any clearer?